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Effectus [21]
2 years ago
12

Of all the Sunny Club members in a particular city, 25% prefer swimming on weekends and 75% prefer swimming on weekdays. It is f

ound that 10% of the members in that city prefer swimming on weekends and are female, while 55% of the members in that city prefer swimming on weekdays and are female. The probability that a club member picked randomly is female, given that the person prefers swimming on weekends, is . NextReset
Mathematics
2 answers:
Semmy [17]2 years ago
6 0

Answer: There is a probability that a club member picked randomly is female, given that the person prefers swimming on weekends is 40%.

Step-by-step explanation:

Let E₁ be the event that members prefer swimming on weekends.

Let E₂ be the event that members prefer swimming on weekdays.

Let A be the event that the members are female.

Probability that members prefer swimming on weekends P(E₁) = 25%

Probability that members prefer swimming on weekdays P(E₂)= 75%

Probability that members prefer swimming on weekends and are female PA∩E₁)= 10%

Probability that members prefer swimming on weekdays and are female P(A∩E₂) = 55%

Using Conditional theorem, we will find the probability that a member is female given that the the person prefers swimming on weekends.

P(A\mid E_1)=\dfrac{P(A\cap E_1)}{P(E_1)}\\\\P(A\mid E_1)=\dfrac{10}{25}\\\\P(A\mid E_1)=0.4\\\\P(A\mid E_1)=0.4\times 100\%=40\%

Hence, there is a probability that a club member picked randomly is female, given that the person prefers swimming on weekends is 40%.

Ivan2 years ago
3 0
P(f | weekend) = p(f & weekend)/p(weekend)
.. = 10%/25%
.. = 2/5 = 0.4
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In figure D we can see that the distance between each point of ΔCED is equal to the distance between each point of ΔMPN. Thus it shows  the triangle pairs which can be mapped to each other using a single translation.


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Which ordered pair would form a proportional relationship with the point graphed below​
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Bruce's vector is <5cos(90-25), 5sin(90-25)> = <5cos(65), 5sin(65)> ≈ <2.1130913087, 4.53153893518>

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Answer:

(1) The degrees of freedom for unequal variance test is (14, 11).

(2) The decision rule for the 0.01 significance level is;

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(3) The value of the test statistic is 0.3796.

Step-by-step explanation:

We are given that you are an expert on the fashion industry and wish to gather information to compare the amount earned per month by models featuring Liz Claiborne's attire with those of Calvin Klein.

The following is the amount ($000) earned per month by a sample of 15 Claiborne models;

$3.5, $5.1, $5.2, $3.6, $5.0, $3.4, $5.3, $6.5, $4.8, $6.3, $5.8, $4.5, $6.3, $4.9, $4.2 .

The following is the amount ($000) earned by a sample of 12 Klein models;

$4.1, $2.5, $1.2, $3.5, $5.1, $2.3, $6.1, $1.2, $1.5, $1.3, $1.8, $2.1.

(1) As we know that for the unequal variance test, we use F-test. The degrees of freedom for the F-test is given by;

\text{F}_(_n__1-1, n_2-1_)

Here, n_1 = sample of 15 Claiborne models

         n_2 = sample of 12 Klein models

So, the degrees of freedom = (n_1-1, n_2-1) = (15 - 1, 12 - 1) = (14, 11)

(2) The decision rule for 0.01 significance level is given by;

  • If the value of our test statistics is less than the critical values of F at 0.01 level of significance, then we have insufficient evidence to reject our null hypothesis.      
  • If the value of our test statistics is more than the critical values of F at 0.01 level of significance, then we have sufficient evidence to reject our null hypothesis.  

(3) The test statistics that will be used here is F-test which is given by;

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where, s_1^{2} = sample variance of the Claiborne models data = \frac{\sum (X_i-\bar X)^{2} }{n_1-1} = 1.007

s_2^{2} = sample variance of the Klein models data = \frac{\sum (X_i-\bar X)^{2} }{n_2-1} = 2.653    

So, the test statistics =  \frac{1.007}{2.653 } \times 1  ~ \text{F}_(_1_4,_1_1_)

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Hence, the value of the test statistic is 0.3796.

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2 years ago
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